A student prepared a stock solution by dissolving 2.50 g of KOH in enough water to make 150. mL of solution. S?
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Answer:
Mol of KOH initially= 2.5/(19+16+1)=0.069444 Mol of KOH in 15ml=15/150*0.069444=0.0069444 =mol in 65ml Hence, conc is 0.0069444/0.065=0.1068mol/L or 3.85g/L
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i got 0.0686 M
Eddie
Mol of KOH initially= 2.5/(19+16+1)=0.069444 Mol of KOH in 15ml=15/150*0.069444=0.0069444 =mol in 65ml Hence, conc is 0.0069444/0.065=0.1068mol/L or 3.85g/L
Neron Austin
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